Hydrodynamic lubrication and Petroff's equation

Hydrodynamic lubrication

When the shaft is at rest it touches the bearing at the bottom. As it begins to rotate, the shaft drags oil into the converging wedge between itself and the bearing because of the viscosity. At sufficient speed the oil in the wedge builds a pressure that lifts the shaft on a full fluid film, with the shaft centre displaced from the bearing centre (eccentricity $e$). The surfaces no longer touch, and the friction is only the viscous shear of the oil.

   bearing
  +-----------+
  |  .--O--.  |   O = journal centre (displaced by e)
  | /  shaft\ |   thin film thickest on the entry side,
  | \_______/ |   thinnest at h_min on the exit side
  +-----------+   pressure peak ahead of h_min

The film thickness and load capacity depend on the Sommerfeld number $S = (\mu N/P)(r/c)^2$, where $\mu$ is the viscosity, $N$ the speed in rev/s, $P = W/(dL)$ the bearing pressure, $r$ the journal radius and $c$ the radial clearance. Higher viscosity and speed and lower load increase the film thickness. The friction coefficient follows the Stribeck curve: high at start (boundary), minimum when the film forms, then slowly rising in the fluid regime.

Petroff's equation

For a concentric journal (light load) the shear stress is $\tau = \mu U/c$ and the friction coefficient is

$$f = 2\pi^2 \frac{\mu N}{P} \frac{r}{c}$$

Calculation

$$\begin{aligned} N &= \frac{1500}{60} = 25\ \text{rev/s} \\ P &= \frac{W}{dL} = \frac{4000}{0.1 \times 0.08} = 5 \times 10^5\ \text{Pa} \\ \frac{r}{c} &= \frac{0.05}{0.00005} = 1000 \\ f &= 2\pi^2 \times \frac{0.025 \times 25}{5 \times 10^5} \times 1000 = 0.02467 \end{aligned}$$ $$\begin{aligned} T &= f W r = 0.02467 \times 4000 \times 0.05 = 4.93\ \text{N m} \\ \text{Power loss} &= 2\pi N T = 2\pi \times 25 \times 4.93 = 775\ \text{W} \end{aligned}$$

Answer: $f = 0.0247$, friction torque $= 4.93$ N m, power loss $\approx 0.775$ kW.